The Timpson Lyceum · General Topology (Point Set Topology)

Continuity & Homeomorphisms – Lecture 15 (Topology)

Overview

This lecture completes the study of continuity and introduces homeomorphisms — the topological notion of two spaces being "the same." Building on the topological definition (the preimage of every open set is open), we prove it is logically equivalent to the calculus $\varepsilon$–$\delta$ definition, show that continuous functions preserve sequence limits, give a closed-set version of continuity, and prove that compositions of continuous functions are continuous (plus the pasting lemma for gluing functions on closed pieces). We then define a homeomorphism as a continuous bijection with continuous inverse, show that "is homeomorphic to" is an equivalence relation, and build intuition with many examples: $\mathbb{R}$ and any open interval are homeomorphic, a circle is not homeomorphic to a half-open interval, and surfaces like a cube and a sphere, or a punctured sphere and the plane, are homeomorphic.

Key Concepts

Intuition for continuity. A topology gives a set the minimum structure needed to express proximity (closeness). A continuous function preserves proximity: points that are close in the domain stay close in the codomain.

Topological definition of continuity. A function $f : X \to Y$ (both topological spaces) is continuous if for every open set $V \subseteq Y$, the preimage $f^{-1}(V)$ is open in $X$. Mnemonic: the preimage of an open set is open.

$\varepsilon$–$\delta$ definition (calculus). $f : \mathbb{R} \to \mathbb{R}$ (standard topology) is continuous at $x_0$ if $\forall\,\varepsilon > 0\ \ \exists\,\delta > 0 \ :\ |x - x_0| < \delta \ \Rightarrow\ |f(x) - f(x_0)| < \varepsilon.$ $f$ is continuous if it is continuous at every point of its domain. (Continuity at a point and continuity of the whole function are distinct notions — one bad point breaks the whole function.)

Neighborhood form (the bridge between the two). $f : X \to Y$ is continuous iff for every $x \in X$ and every open set $V \ni f(x)$, there exists a neighborhood $U_x \ni x$ with $f(U_x) \subseteq V$. Picking an $\varepsilon$ corresponds to picking an open set $V$ around $f(x)$; finding a $\delta$ corresponds to finding a neighborhood $U_x$ around $x$.

Continuity and sequences. If $f : X \to Y$ is continuous and $x_n \to x_0$ in $X$, then $f(x_n) \to f(x_0)$ in $Y$. Continuous functions carry convergent sequences to convergent sequences — proximity is preserved, and limits are exactly about proximity. (Recall $x_n \to x_0$ means: for every neighborhood $U \ni x_0$ there is $N \in \mathbb{N}$ with $x_n \in U$ for all $n \ge N$.)

Closed-set characterization. $f : X \to Y$ is continuous iff for every closed set $D \subseteq Y$, the preimage $f^{-1}(D)$ is closed in $X$. (Proof uses $f^{-1}(Y \setminus D) = X \setminus f^{-1}(D)$, the fact that complements of open sets are closed, and vice versa.)

Composition. If $f : X \to Y$ and $g : Y \to Z$ are continuous, then $g \circ f : X \to Z$ is continuous, since $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$ is the preimage of an open set under $f$ of an open set under $g$. Combined with "polynomials and $|x|$ are continuous," this instantly certifies many functions, e.g. $\left|\,3x^3 - 27x + 15x^2\,\right|$ is continuous.

Pasting lemma. Let $A, B$ be closed in $X$ with $A \cup B = X$, and let $f : A \to Y$, $g : B \to Y$ be continuous with $f(x) = g(x)$ for all $x \in A \cap B$. Then the glued function $h(x) = \begin{cases} f(x) & x \in A \\ g(x) & x \in B \end{cases}$ is continuous. Example: $|x|$ is continuous because it pastes $-x$ (on $(-\infty,0]$) and $x$ (on $[0,\infty)$), which agree at $x = 0$.

Homeomorphism. Let $X, Y$ be topological spaces and $f : X \to Y$ a bijection. $f$ is a homeomorphism if both $f$ and $f^{-1}$ are continuous. If such an $f$ exists, $X$ and $Y$ are homeomorphic, written $X \cong Y$. Intuitively, $f$ pairs off points and open sets one-to-one, preserving proximity in both directions. A bijection alone says the underlying sets have the same cardinality ("same size"); continuity both ways adds preservation of topological structure.

$\cong$ is an equivalence relation.

• Reflexive: $X \cong X$ via the identity map.
• Symmetric: if $f : X \to Y$ is a homeomorphism, so is $f^{-1} : Y \to X$.
• Transitive: if $f : X \to Y$ and $g : Y \to Z$ are homeomorphisms, so is $g \circ f$ (composition of bijections is a bijection; composition of continuous maps is continuous, and likewise for the inverses).

Catalog of homeomorphic spaces (standard topology).

• $\mathbb{R} \cong (-1,1)$ via $f(x) = \dfrac{x}{1 + |x|}$ (squashes the whole line into a bounded interval).
• $(-1,1) \cong (a,b)$ for any bounded open interval, hence $\mathbb{R} \cong (a,b)$ by transitivity.
• $\mathbb{R} \cong (a, \infty)$ via $f(x) = e^{x} + a$ (always $> a$, never reaching $a$).
• All bounded closed intervals are homeomorphic to each other (but not to $\mathbb{R}$ — you would have to "map $\pm\infty$" to the endpoints).
• All half-open intervals are homeomorphic; e.g. $(a, \infty) \cong (-\infty, b)$ via $h(x) = a + b - x$.
• A circle $S^1$ is not homeomorphic to $[0, 360)$: the parametrization is a continuous bijection, but its inverse is not continuous — points near $0$ and near $360$ are close on the circle yet far apart on the interval.
• $\mathbb{R}^2 \cong$ an open half-plane, and $\cong$ an open disk (via $r \mapsto \tfrac{r}{1+r}$ in polar coordinates); a square and a circle (disk) are homeomorphic.
• The surface of a cube $\cong$ the surface of a sphere (radial projection).
• A punctured sphere $\cong \mathbb{R}^2$ (stereographic projection from the missing point) — removing one point lets you "stretch the hole" flat.

Stretching, shrinking, and the "rubber sheet" caveat. Shifting, stretching, and compressing never change topological type (distances are not preserved, but proximity is). The rubber-sheet picture is a helpful but imperfect guide: a cut-and-reglue that keeps all nearby points nearby still yields a homeomorphism, so a once-extra-twisted band can be homeomorphic to a plain cylinder. Finer distinctions (e.g. detecting the twist) need a stronger notion of sameness introduced later.

Worked Example

Claim. $\mathbb{R}$ with the standard topology is homeomorphic to the open interval $(-1, 1)$, via $f(x) = \dfrac{x}{1 + |x|}$.

Step 1 — values and behavior. Tabulate the map and note its limiting behavior:

Step 2 — bijection. $f$ is strictly increasing, so distinct inputs give distinct outputs (one-to-one). Its outputs fill all of $(-1,1)$ and never escape it, so it is onto $(-1,1)$. Hence $f$ is a bijection onto $(-1,1)$.

Step 3 — continuity both ways. The denominator $1 + |x| \ge 1$ is never zero, so $f$ is a quotient of continuous functions with nonvanishing denominator and is continuous; equivalently, you never lift your pencil. The inverse is likewise continuous. Therefore $f$ is a homeomorphism and $\mathbb{R} \cong (-1,1)$.

Representative proof — continuity $\Rightarrow$ neighborhood form. Assume $f : X \to Y$ is continuous, fix $x \in X$, and let $V \subseteq Y$ be open with $f(x) \in V$. Set $U_x = f^{-1}(V)$. Since $f$ is continuous, $U_x$ is open in $X$; and $x \in U_x$ because $f(x) \in V$. Then $f(U_x) = f\!\big(f^{-1}(V)\big) \subseteq V$. So a neighborhood $U_x$ with $f(U_x) \subseteq V$ always exists. $\blacksquare$

Summary

• A map $f : X \to Y$ is continuous iff the preimage of every open set is open — equivalently, iff the preimage of every closed set is closed, and equivalently (on metric/standard spaces) the $\varepsilon$–$\delta$ condition holds.
• Continuity preserves proximity: continuous maps send convergent sequences $x_n \to x_0$ to convergent sequences $f(x_n) \to f(x_0)$.
• Compositions of continuous functions are continuous, and the pasting lemma glues functions defined continuously on two closed pieces that agree on the overlap.
• A homeomorphism is a continuous bijection with continuous inverse; "$\cong$" is an equivalence relation (reflexive, symmetric, transitive) capturing when two spaces are topologically the same.
• $\mathbb{R}$ is homeomorphic to every bounded open interval, to half-infinite open rays, and to open half-planes/disks; closed bounded intervals are homeomorphic to each other but not to $\mathbb{R}$, and a circle is not homeomorphic to a half-open interval.
• To show two spaces are homeomorphic, exhibit one homeomorphism; to show they are not, you must rule out all such maps — far harder, later done via topological properties that homeomorphisms preserve.